I want to compare two Feynman diagrams and be able to say which one describes a process that is more likely to happen.

As far as I understand, this is done by considering the order of the diagram.

In the case of, for example, $e^+ e^- \rightarrow \mu^+ \mu^-$ (tree level) I have the two electrons coupling to the photon - introducing a factor of $\alpha_{EM}$, the coupling constant - and the two muons coupling to the same photon, introducing another factor of $\alpha_{EM}$.

At tree level, therefore, $$\Gamma_{tree} (e^+ e^- \rightarrow \mu^+ \mu^-) \propto \alpha_{EM}^2,$$ but if I were to include an electron-positron loop I'd have another $\alpha_{EM}^2$ therefore giving $$\Gamma_{1 loop} (e^+ e^- \rightarrow \mu^+ \mu^-) \propto \alpha_{EM}^4$$ and so on. In this case I understand that the power of $\alpha_{EM}$ is the order of the diagram.

ANYWAY, what I really want to quantify is this kind of processes:

enter image description here

(a) I have 4 vertices of the $W$ bosons... so what, is this $\propto \alpha_{weak}^4$?

(b) By the same logic, $\propto \alpha_{EM}*\alpha_{weak}^2$?

Is (a) less probable than (b) then?

  • $\begingroup$ You can't say that for sure unless you actually calculate the diagrams. $\endgroup$ – ACuriousMind Apr 14 '15 at 17:45
  • $\begingroup$ So there's no way of knowing any quantitative information from the diagram alone? $\endgroup$ – SuperCiocia Apr 14 '15 at 17:46
  • $\begingroup$ The diagram is just shorthand for (a part of) the integrals you have to compute to get (one term in the perturbation series for) the probabilities. $\endgroup$ – ACuriousMind Apr 14 '15 at 17:47
  • $\begingroup$ OK. In the literature it always says that diagrams (a) and (b) represent the leading order diagram for this process (flavour changing neutral current). Does it mean that they did the calculation? Or is there a quick way of seeing they have the same order? $\endgroup$ – SuperCiocia Apr 15 '15 at 11:13
  • $\begingroup$ My naive idea was that, since (a) had more vertices, it would be less probable. Since each vertex contributes a coupling constant, which is < 1 $\endgroup$ – SuperCiocia Apr 15 '15 at 11:14

Your reasoning is essentially correct. Assuming that $\alpha_\text{weak}^4 \ll \alpha_\text{EM}\alpha_\text{weak}^2$, we can conclude that (a) is less probable than (b). As ACuriousMind notes, to be certain one should actually compute the full diagram, but the integrations from the loops usually give quite reasonable numbers that don't change the picture much, except maybe for specific values of external paramters. For example, a diagram like (a) may dominate if the virtual quarks are on-shell (although I don't think it'll matter in this particular case).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.